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Q.
Which one of the following is the rationalize form of $\frac{11}{\sqrt{17}-\sqrt{6}}$ ?
Real Numbers
Solution:
$\frac{11}{\sqrt{17}-\sqrt{6}} \times \frac{\sqrt{17}+\sqrt{6}}{\sqrt{17}+\sqrt{6}}$
Using $(a+b)(a-b)=a^2-b^2$, we get
$\frac{11(\sqrt{17}+\sqrt{6})}{(\sqrt{17})^2-(\sqrt{6})^2}=\frac{11(\sqrt{17}+\sqrt{6})}{17-6}
=\frac{11(\sqrt{17}+\sqrt{6})}{11} $
$=\sqrt{17}+\sqrt{6}$